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**Shahana****Member**- Registered: 2012-08-17
- Posts: 3

Prove that for all real numbers x>=1,

3|x-2|/x <= 4

My attempt:

I took the |x-2| to be (x-2). I think this is where i am making a mistake, regarding the modulus sigh.

3(x-2)/x <= 4

3(x-2) <= 4x

3x - 6 <= 4x

3x - 4x - 6 <= 0

-x - 6 <= 0

x >= -6

Please help me!

Thank you.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi Shahana;

The attempt to prove x>=1 will be difficult. What about x = - 1?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Shahana****Member**- Registered: 2012-08-17
- Posts: 3

Hi bobbym,

No i am supposed to prove x>=1. I cannot understand how to do it.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

You can not prove that x>=1.

It must be a given and then you are trying to prove the inequality.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Shahana****Member**- Registered: 2012-08-17
- Posts: 3

Yes in fact i am supposed to prove the inequality for all real numbers x>=1

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**daydoe12****Member**- Registered: 2012-11-18
- Posts: 8

bobbym how would you figure out the answer to this question because I am now having problems with this

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

Did you click the hidden text in post #6?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**daydoe12****Member**- Registered: 2012-11-18
- Posts: 8

no but i will thanks

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

See if you can use it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**daydoe12****Member**- Registered: 2012-11-18
- Posts: 8

how is that proving for all real numbers

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

It proves it for all real numbers greater then 6 / 7.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**daydoe12****Member**- Registered: 2012-11-18
- Posts: 8

ok that makes sense

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

I hope it is what the OP wanted, but he never came back. I do not know if that is good or bad.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

bobbym wrote:

Hi;

There is a minor flaw in this proof. You are using the second case for all x>=1, when you should be using it for x in [2,infty).

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

Why two? We are not talking about integers.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

The first case is for reals in [1,2) because that is where x-2 is positive. Only for larger numbers should the second case be used.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

I am not following you. Is it not true that it is true for x >= 6 / 7? Does that not imply greater than 1 also?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

No, no. The inequality comes down to x>=6/7 only for x>=2. To prove it for 1<=x<2 you have to use the case where you get x>=-6.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

That I am not getting at all. Care to explain it?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

First we need to split the problem into two cases.

The first case is when x is in [1,2). In that case the equation becomes 3*(x-2)/x<=4 which then comes down to x>=-6. Since this is true for x in the said interval, we can move on to the second case, which is x>=2. In this case the equation becomes 3*(2-x)/x<=4 and reduces to x>=6/7, which is true for x>=2. Since both cases are satisfied the proposition holds.

Here lies the reader who will never open this book. He is forever dead.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi!

Those of you who have access to a good graphing program might try graphing the function

f(x) = 3|x-2|/x and then draw the horizontal line y=4 and see where the graph lies below

or on that line. I think you may find that any x in (-inf, 0) union [6/7, inf) will have its output

less than or equal to 4. (I graphed it by hand so that could be a "bit shaky". I'd like to see

the graph of this function using a good graphing program. Hint! Hint!)

I'm still a bit confused about what the original question is. The domain of the function above

is any real number except zero. The range seems to be the set R-[-3, 0), R being the set of reals.

If we require f(x) <= 4 then this is true for a subset of the domain, namely that mentioned above:

(-inf,0) union [6/7, inf). Any x in the subset [1, inf) of the domain will certainly satisfy f(x) <= 4.

If we require that f(x) be in [-3, 0) then the domain is restricted to the empty set.

Any restriction on the outputs of f(x) corresponds to a subset of the domain R-{0}.

Solving the original inequality is essentially finding out what subset of the domain of f(x) will make

the inequality true. To do so I usually consider all possible cases that make a difference in the

expressions replacing the absolute value expressions. It's a bit tedious, but I tend to get lost if I

don't consider each case. The union of the sets obtained in each case is typically the final answer.

Also in each case it is the intersection of the assumed condition and its result that gives the set for

that case.

Determining the range of the function is a different matter altogether.

Note: We could also make the function f(x) = (3|x-2|/x) - 4 and see where it is <= zero.

Absolute value inequalities are tricky. It is probably best to write an appropriate function f(x)

and graph it to see what the solutions are keeping in mind that the solution set is a subset of

the domain of the function.

GOTTA GET SOME

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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